我的 AWS s3 存储桶中有多个格式文件,例如 pdf、doc、rtf、odt、png,我需要从中提取文本。我已经设法通过它们的路径获取内容列表。现在根据文件类型,我将使用不同的库从文件中提取文本。由于文件可能有数千个,我需要直接从 s3 中提取文本,而不是下载。
filespath=['https://abc.s3.ap-south-1.amazonaws.com/DocumentOnPATest', 'https://abc.s3.ap-south-1.amazonaws.com/IndustryReport2019.pdf', 'https://abc.s3.ap-south-1.amazonaws.com/receipt.png', 'https://abc.s3.ap-south-1.amazonaws.com/sample.rtf', 'https://abc.s3.ap-south-1.amazonaws.com/sample1.odt']
bucketname =abc
我尝试了一些东西,但它给了我错误
for path in filespath:
ext=pathlib.Path(path).suffix
if ext=='.pdf':
pdf_file=PyPDF2.PdfFileReader(path)
print(pdf_file.extractText())
但我收到一个错误
File "F:\Projects\FileExtractor\fileextracts3.py", line 28, in <module>
pdf_file=PyPDF2.PdfFileReader(path)
File "C:\ProgramData\Anaconda3\lib\site-packages\PyPDF2\pdf.py", line 1081, in __init__
fileobj = open(stream, 'rb')
OSError: [Errno 22] Invalid argument: 'https://abc.s3.ap-south-1.amazonaws.com/IndustryReport2019.pdf
请帮我领导。谢谢